Leetcode 301. Remove Invalid Parentheses

Remove the minimum number of invalid parentheses in order to make the input string valid. Return all possible results.

Note: The input string may contain letters other than the parentheses ( and ).

Examples:

"()())()" -> ["()()()", "(())()"]"(a)())()" -> ["(a)()()", "(a())()"]")(" -> [""]

Credits:
Analysis: this question is quite similar to all the Parentheses question. we need to make sure that the left ( is more than ), then if the number of the left ( is equal to the number of the right one ), there is a valid parentheses. then we just compare is to the maxValue.

1 This problem has a trick that how to delete the duplicates. this is quite common in backtracking

if(i != start && s.charAt(i)== s.charAt(i - 1))
          continue;

2 also consider the "", when the input string is "", we need to return a "" in the result

code 

public class Solution {    int maxNum = 0;    public List<String> removeInvalidParentheses(String s) {      int left = 0;      int right = 0;            List<String> resultList = new ArrayList<String>();      resultList.add("");      int length = s.length();      if(length == 0){        return resultList;      }      int start = 0; //start is to record the begining bit to visit the String      dfs( s, "", start, left, right,resultList,0);      return resultList;    }    public void dfs(String s, String result, int start, int left, int right,List<String> resultList, int number){      if(left == right && number != 0){        if(number > maxNum){          maxNum = number;          resultList.clear();          resultList.add(result);        }        else if(number == maxNum){          resultList.add(result);        }      }      int length = s.length();      if(start >= length)        return;      for(int i = start; i < length; i++){        if(i != start && s.charAt(i)== s.charAt(i - 1))          continue;        if(s.charAt(i) == '(')          dfs(s,result + "(", i + 1, left + 1, right,resultList,number + 1);        else if(s.charAt(i) == ')' && left > right)          dfs(s,result + ")", i + 1, left, right + 1,resultList,number + 1);        else if(s.charAt(i) != '(' && s.charAt(i) != ')')          dfs(s,result + s.charAt(i),i + 1,left,right,resultList, number + 1);      }    }}